Proving that the Lindeberg condition Fails I am having trouble proving the following result thanks for any help in advance:

Let {$X_{n}, n\geq 1$} be a sequence of independent random variables with $P(X_{n} = -n) = 1-1/n^{2}$ and $P(X_{n} = n^{3} - n) = 1/n^{2}, n \geq 1$ Prove that {$X_{n}, n\geq1$} does not obey the Lindeberg condition.

There are a few things that I have noticed:
1.$\sum_{j = 1}^{n} X_{j} / n \rightarrow -\infty$ almost certainly
2.$EX_{n}^{2} /\sum_{j = 1}^{n}EX_{j}^{2}\rightarrow 0$ and $\sum_{j = 1}^{n}EX_{j}^{2} \rightarrow \infty$, so if I assume the Lindeberg condition is satisfied then by the Lindeberg-Feller CLT there will be some contradiction stemming from $\sum_{j = 1}^{n} X_{j}/\sqrt(\sum_{j = 1}^{n}EX_{j}^{2})$ converging in distribution to a $\mathcal{N} (0,1)$ random variable.


*I imagine that I am trying to get some contradiction  with 1.

 A: Observe that $X_n$ is centered and that for all $n$, 
\begin{align}
\mathbb E\left[X_n^2\right]&=n^2\left(1-n^{-2}\right)+\left(n^3-n\right)^2n^{-2}\\
&= n^2-1+n^2\left(n^2-1\right)^2n^{-2}\\
&= \left(n^2-1\right)\left(1+n^2-1\right)
\\
&= n^2\left(n^2-1\right)
\end{align}
hence there exists constants $c$ and $C$ such that 
$$
cn^5\leqslant\sum_{i=1}^n\operatorname{Var}\left(X_i\right)\leqslant Cn^5
$$
and the Lindeberg condition is equivalent to the following:
$$
\forall \varepsilon\gt 0, \frac 1{n^5}\sum_{i=1}^n\mathbb E\left[X_i^2\mathbf 1\left\{\left\lvert X_i\right\rvert>\varepsilon n^{5/2}\right\}\right]=0.
$$
We will show that this fails for $\varepsilon=1$. Let us denote by $I_n$ the set of integers $i$ such that $2\leqslant i\leqslant n$ and $i^3-i\gt n^{5/2}$. Then 
\begin{align}
\frac 1{n^5}\sum_{i=1}^n\mathbb E\left[X_i^2\mathbf 1\left\{\left\lvert X_i\right\rvert> n^{5/2}\right\}\right]&\geqslant \frac 1{n^5}\sum_{i\in I_n}\mathbb E\left[X_i^2\mathbf 1\left\{\left\lvert X_i\right\rvert>  n^{5/2}\right\}\right]\\
&\geqslant \frac 1{n^5}\sum_{i\in I_n}\mathbb E\left[X_i^2\mathbf 1\left\{  X_i=i^3-i\right\}\right]\\
&=  \frac 1{n^5}\sum_{i\in I_n}\left(i^3-i\right)^2i^{-2}\\
&\geqslant \frac 1{n^5}\sum_{i\in I_n}\left(i^2-1\right)^2.
\end{align}
Observe that for $n$ large enough event, $I_n$ contains the set of integers between $n/2$ and $n$ hence 
$$
\frac 1{n^5}\sum_{i=1}^n\mathbb E\left[X_i^2\mathbf 1\left\{\left\lvert X_i\right\rvert> n^{5/2}\right\}\right] \geqslant \frac 1{n^5}\sum_{i=n/2}^n\left(i^2-1\right)^2\geqslant \frac 1{2n^4}\left(\left(\frac n2\right)^2-1\right)^2.
$$
